working on an Arduino project powered from an "auto-shutting OFF" battery pack, I learned about oscillator ckrcuits used to draw power from a battery pack momentarily in order to keep it from shutting off.

And I'm trying to understand how it works. I understand how a capacitors, inductors and bjt's work. But how does an oscillator circuit put these elements together as in this example, to make an led blink?

This LTspice simulation below shows that the circuit does not appear to operate as advertised.
The current through the 22 ohm resistor is a sawtooth, not a short pulse.

You would need a different circuit to get a short pulse.
For example, below is a PWM circuit using CD40106 hex Schmitt trigger inverter gates. The PWM duty-cycle is adjusted such that it generates a ≈20ms pulse every 1.3s.

The six gates in a package are connected in parallel to give a higher output drive current.

It uses very little power other than that from the periodic 220mA pulses.

The circuit does work but it want to be as in the arrangement of circuit #2:

The circuit starts by the 330k turning ON the PNP transistor very slightly. The 10u is uncharged.
The current through the collector-emitter of the PNP transistor turns ON the NPN transistor a small amount.
This brings the right-hand lead of the 10u towards the 0v rail.
The left lead of the 10u tries to fall also but it is restrained by the base of the PNP transistor.
Pulling the leads apart like this does one thing. It allows the 10u to charge and this causes current to flow through the 10u and the emitter-base of the PNP transistor.
This causes the PNP transistor to turn ON much harder and the current through the c-e leads also turns on the NPN transistor very hard.
This puts a very high current through the 22R and also the LED in the emitter of the transistor to produce a very bright flash.
The 10u charges very quickly because there is virtually no current limiting resistors in the charging circuit.
As it gets to the point of being fully charged, the current flowing through it decreases to a very small amount and this current is flowing in the emitter-base circuit of the PNP transistor.
It gets to a point that the transistor starts to turn OFF and the current through the 22R decreases so that the voltage across the 22R reduces and the right lead of the 10u rises towards the 5v rail. The left lead follows and the voltage on the base of the PNP transistor actually rises above the 5v rail and as soon as it rises to the value of 4.4v, the PNP transistor completely turns off and the LED is off for a very long time.
The only component in the circuit that will discharge the 10u is the 330k and it does this very slowly and then it starts to charge the 10u to start the next cycle.

OK Im beginning to see how colin555 circuit works. I see that he and Dana added a 1k & 10nF cap in parallel and I'll get to that later.
So when powered on, current flows thru the 330k to the base if the pnp, which turns on the npn a little as well?

IS it because the 22 ohm R is the path of least resistance so it flows thru the cap and to the pnp base?

So when powered on, current flows thru the 330k to the base if the pnp, which turns on the npn a little as well?

Click to expand...

That's during startup.
But the interesting operation is during the steady-state pulsing so I'll attempt to describe that with the aid of the simulation waveforms below.
As can be seen, during the off portion of the cycle, the PNP base voltage (yellow trace) starts from near double the 6V supply voltage (about 11V) and falls with a time-constant of R3C1.

When the PNP base emitter voltage (red trace) reaches about -0.6V at the end of the fall it starts to turn on.
This also turns on the NPN immediately, pulling its collector voltage (blue trace) to zero
This rapidly discharges C1 with an R2C1 time-constant through the base of Q2, and when Q2's base current drops below its turn on value, Q2 starts turning off.
This turns off Q1, causing the right side of cap capacitor to rapidly rise to 6V and the left side of the capacitor to also rise 6V from about 5.3V to 11V (Q1 Vbe to about -5V),

I don't want to use the Arduino to keep the battery pack because I want to save power to keep the battery pack as charged as possible.

I still don't understand what makes the led blink. My first doubt is, why doesn't the current flow through the 22 ohm resistor first and thru the cap, instead of through the 330k and to the base of the 3906?